Introduction to Quadratics: Additional Practice (Set 1)

Take the square root of both sides, keeping both signs.

$x - 7 = \pm 2$

Then $x = 7 + 2 = 9$ or $x = 7 - 2 = 5$.

Set each factor equal to $0$ and solve.

$4x + 8 = 0 \;\Rightarrow\; x = -2, \qquad x - 6 = 0 \;\Rightarrow\; x = 6$

So $x = -2$ or $x = 6$.

Square the binomial by multiplying it by itself.

$(x - 6)^2 = (x - 6)(x - 6) = x^2 - 6x - 6x + 36 = x^2 - 12x + 36$

Since $100 > 0$, two real numbers square to it.

$x = \pm 10$

So there are two real solutions.

Isolate the squared term by subtracting $5$ from both sides.

$x^2 = -4$

No real number squares to a negative, so there is no real solution.

The factors are $3x$ and $x - 4$. Set each to $0$; note $3x = 0$ still gives $x = 0$.

$x = 0 \quad \text{or} \quad x = 4$

The constant $3$ does not create a root, but do not lose $x = 0$.

Move every term to the left so the right side becomes $0$.

$2x^2 + 6 = 7x \;\Rightarrow\; 2x^2 - 7x + 6 = 0$

So $a = 2$, $b = -7$, $c = 6$; the coefficient $b = -7$.

Take the square root of both sides, keeping both signs.

$x + 5 = \pm 6$

Then $x = -5 + 6 = 1$ or $x = -5 - 6 = -11$.

Substitute $x = -4$ into $x^2 - 16$.

$(-4)^2 - 16 = 16 - 16 = 0$

So $-4$ is a solution. In fact $x = 4$ is a solution too, so it is not the only one; a negative value can absolutely be a root.

Multiply each term of the first factor by each term of the second, then combine like terms.

$(3x + 1)(x - 2) = 3x^2 - 6x + x - 2 = 3x^2 - 5x - 2$

Take the square root of both sides and simplify $\sqrt{20} = 2\sqrt{5}$.

$x = \pm\sqrt{20} = \pm 2\sqrt{5}$

Both signs are kept, giving two irrational roots.

Both factors are the same, so the only way the product is $0$ is $x - 2 = 0$.

$x - 2 = 0 \;\Rightarrow\; x = 2$

The root $x = 2$ is repeated, so there is just one distinct solution.